An object moving in a circle of radius r with constant speed v has acceleration of magnitude a_c = v² / r directed toward the centre of the circle (centripetal). In terms of angular speed ω = v/r, a_c = ω² r. The acceleration changes direction continuously even though the speed is constant.
-- NCERT Class 11 Physics, Ch. 3, p. 42Uniform Circular Motion
8 MCQs2 revision cards9-step worked example
Source: NCERT KinematicsPYQ coverage: NEET 2024Official key: NTA-verifiedLast updated: 25 Sep 2026
Uniform Circular Motion, explained for NEET
The trap: aspirants see "constant speed" in uniform circular motion (UCM) and conclude there is zero acceleration. NEET exploits this confusion reliably — if you mark "acceleration = 0" because the speedometer doesn't change, you lose marks.
What UCM actually is. A particle moves along a circle of radius r at constant speed v. The speed (scalar) never changes. But velocity is a vector — it has both magnitude and direction. Because the direction of motion rotates continuously along the circle, the velocity vector changes every instant, which means there is a non-zero acceleration (NCERT Class 11 Physics Chapter 3, page 42).
Centripetal acceleration. This acceleration points radially inward — toward the centre of the circle — and has magnitude:
$$a_c = \frac{v^2}{r} = \omega^2 r$$
where ω = v/r is the angular speed. The word "centripetal" means "centre-seeking." There is no tangential acceleration component in UCM because the speed is constant — the entire acceleration is radial.
What stays constant, what doesn't.
| Quantity | Constant in UCM? | Why |
|---|---|---|
| Speed | Yes | Definition of "uniform" |
| Kinetic energy (½mv²) | Yes | Depends only on speed |
| Velocity | No | Direction changes continuously |
| Acceleration magnitude | Yes | v²/r with constant v, r |
| Acceleration direction | No | Always points toward centre; rotates with the particle |
The UCM-to-projectile bridge. A common NEET pattern gives you a particle in UCM (radius R, period T) and then asks: "if this particle is now launched vertically upward with the same speed, find the maximum height." The bridge step is computing v = 2πR/T from the circular motion data before applying projectile formulas. Do not set H = R — the radius and the projectile height are unrelated quantities.
Can you answer these Uniform Circular Motion MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
A particle moves in a circle of radius 0.50 m at a constant speed of 4.0 m/s. What is the magnitude of its centripetal acceleration?
Show answer and why every option is right or wrong
Answer: B. a_c = v²/r = (4.0)²/0.50 = 16/0.50 = 32 m/s² (NCERT Class 11 Physics Chapter 3, page 42).
Why A is wrong: A is wrong because 2.0 m/s² comes from multiplying v by r (4.0 × 0.50) instead of dividing v² by r.
Why C is wrong: C is wrong because 16 m/s² results from v² = 16 but forgets to divide by r = 0.50 m (equivalently, divides by 1 instead of 0.50).
Why D is wrong: D is wrong because 8.0 m/s² comes from v/r instead of v²/r. The formula requires squaring the speed.
MCQ 2Easy RecallPYQ Pattern
In uniform circular motion at constant speed, which of the following quantities is NOT constant?
Show answer and why every option is right or wrong
Answer: B. Velocity is a vector. In UCM the direction of motion changes continuously, so velocity is not constant even though its magnitude (speed) is (NCERT Class 11 Physics Chapter 3, page 42).
Why A is wrong: A is wrong because KE = ½mv²; since speed v is constant in UCM, kinetic energy remains constant (trap: confusing velocity change with energy change).
Why C is wrong: C is wrong because constant speed is the defining feature of uniform circular motion.
Why D is wrong: D is wrong because |a_c| = v²/r; both v and r are constant in UCM, so the acceleration magnitude is constant. Only its direction changes.
MCQ 3Concept TrapPractice
A body moves in a circle at constant speed. A student claims: 'Since the speed is constant, the acceleration is zero.' Which statement correctly identifies the error?
Show answer and why every option is right or wrong
Answer: A. Acceleration is the rate of change of velocity (a vector). In UCM, speed is constant but direction changes, so velocity changes and acceleration is non-zero — it equals v²/r directed toward the centre (NCERT Class 11 Physics Chapter 3, page 42).
Why B is wrong: B is wrong because it repeats the student's error. Acceleration depends on the rate of change of velocity (vector), not speed (scalar). Direction change produces centripetal acceleration (trap: UCM_VELOCITY_VS_SPEED).
Why C is wrong: C is wrong because speed IS constant in uniform circular motion — that's what 'uniform' means. The error lies in conflating speed with velocity, not in speed varying.
Why D is wrong: D is wrong because kinetic energy (½mv²) is actually constant in UCM since speed is constant. The existence of acceleration in UCM comes from the changing direction of velocity, not from any energy change.
MCQ 4Easy RecallPractice
The centripetal acceleration of a particle in uniform circular motion is directed:
Show answer and why every option is right or wrong
Answer: C. Centripetal means 'centre-seeking.' The acceleration in UCM always points radially inward toward the centre of the circle (NCERT Class 11 Physics Chapter 3, page 42).
Why A is wrong: A is wrong because a tangential acceleration would change the speed. In UCM speed is constant, so there is no tangential component — the entire acceleration is radial.
Why B is wrong: B is wrong for the same reason as A: tangential direction (whether forward or backward) would alter the speed, contradicting the 'uniform' condition.
Why D is wrong: D is wrong because 'radially outward' describes centrifugal — a pseudo-force in a rotating frame. In the inertial frame, the acceleration is centripetal (inward).
MCQ 5Direct ApplicationPractice
A particle in uniform circular motion has angular speed ω = 4.0 rad/s and moves on a circle of radius 0.25 m. What is the centripetal acceleration?
Show answer and why every option is right or wrong
Answer: D. a_c = ω²r = (4.0)² × 0.25 = 16 × 0.25 = 4.0 m/s² (NCERT Class 11 Physics Chapter 3, page 42).
Why A is wrong: A is wrong because 1.0 m/s² results from computing ωr = 4.0 × 0.25 = 1.0 — this gives the tangential speed v, not the acceleration. The formula requires ω², not ω.
Why B is wrong: B is wrong because 2.0 m/s² comes from halving ω²r = 4.0, as if the ½ of ½mv² belonged in the centripetal formula. a_c = ω²r has no factor ½.
Why C is wrong: C is wrong because 16 m/s² comes from ω² = 16 alone, forgetting to multiply by r = 0.25 m.
MCQ 6Easy RecallPractice
In uniform circular motion, the work done by the centripetal force over one complete revolution is:
Show answer and why every option is right or wrong
Answer: C. The centripetal force is always perpendicular to the velocity (radially inward while motion is tangential). Work = F · d cos θ; with θ = 90° at every instant, cos 90° = 0, so the work done is zero.
Why A is wrong: A is wrong because it treats the centripetal force as if it acts along the displacement over the full circumference. Since force is perpendicular to velocity at every point, no work is done regardless of path length.
Why B is wrong: B is wrong — mv² has dimensions of energy but the centripetal force does no work because it is always perpendicular to the instantaneous displacement.
Why D is wrong: D is wrong for the same reason: centripetal force is perpendicular to displacement at every instant, so the dot product F·ds = 0 throughout the motion.
MCQ 7CalculationPYQ Pattern
A particle moves in a circle of radius R with period T. If this particle is then launched vertically upward with the same speed it had in the circular motion, the maximum height reached is:
Show answer and why every option is right or wrong
Answer: D. Step 1: UCM speed v = 2πR/T. Step 2: For vertical launch, H = v²/(2g) = (2πR/T)²/(2g) = 4π²R²/(2gT²) = 2π²R²/(gT²). This is the UCM-to-projectile bridge calculation.
Why A is wrong: A is wrong because it equates the circular radius with the projectile height. The radius R and the height H are physically unrelated quantities — H depends on v² and g (trap: UCM_TO_PROJECTILE_SPEED_BRIDGE).
Why B is wrong: B is wrong because 2πR/T has dimensions of speed (m/s), not height (m). This is the UCM speed v, not the max height H = v²/(2g).
Why C is wrong: C is wrong because gT²/(4π²) has dimensions of length but corresponds to R itself (from v = 2πR/T → R = vT/(2π) → R = gT²/(4π²) only if v = gT/(2π), which is not given). It is not the max height formula.
MCQ 8Direct ApplicationPractice
A stone tied to a string moves in a horizontal circle of radius 1.0 m at constant speed. If the magnitude of centripetal acceleration is 25 m/s², the speed of the stone is:
Show answer and why every option is right or wrong
Answer: A. a_c = v²/r → v² = a_c × r = 25 × 1.0 = 25 → v = 5.0 m/s (NCERT Class 11 Physics Chapter 3, page 42).
Why B is wrong: B is wrong because 2.5 m/s is half the correct speed, as if a factor ½ belonged in the relation. The correct relation gives v = √(a_c · r) = √25 = 5.0 m/s.
Why C is wrong: C is wrong because 12.5 m/s comes from a_c × r / 2 = 12.5, then failing to take the square root. The formula is v = √(a_c · r), not a_c · r / 2.
Why D is wrong: D is wrong because 25 m/s treats a_c = v directly (a_c = v/r → v = a_c × r = 25). The actual formula is a_c = v²/r, so v = √(a_c × r), not a_c × r.
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Uniform Circular Motion: quick recall before you leave
How do you solve a Uniform Circular Motion question? A worked example
Pattern: UCM-to-projectile bridge (NEET pattern: projectile from circular motion, anchored to 2021 P3 Q49)
- 1
Given
• Radius R = 0.50 m• Period T = 0.20 s• g = 10 m/s² (exact, problem-defined)
- 2
Required
Maximum height H when launched vertically upward with speed v from UCM.
- 3
Concept
In UCM, the tangential speed is v = 2πR/T. When launched vertically, the particle undergoes uniformly decelerated motion under gravity. At maximum height, final velocity = 0. Use the kinematic relation: H = v²/(2g).
- 4
Formula
$$v = \frac{2\pi R}{T}, \qquad H = \frac{v^2}{2g}$$
- 5
Substitution
$$v = \frac{2\pi \times 0.50}{0.20} = \frac{\pi}{0.20} = 5\pi \text{ m/s}$$
$$H = \frac{(5\pi)^2}{2 \times 10} = \frac{25\pi^2}{20}$$ - 6
Calculation
$$H = \frac{25 \times 9.87}{20} = \frac{246.7}{20} = 12.3 \text{ m}$$
(Using π² ≈ 9.87.)
Note on exact constants: g = 10 m/s² is stated as exact in the problem. The integers 2 in the denominator and the factor 2π are mathematical constants. These do not limit significant figures. The measured quantities (R = 0.50 m, T = 0.20 s) each have 2 significant figures, so the answer is reported to 2 significant figures: H ≈ 12 m (or 2π²R²/(gT²) = 12.3 m if left in exact form). - 7
Final answer
$$H \approx 12 \text{ m}$$
Or in exact closed form: H = 2π²R²/(gT²). - 8
Common trap
Setting H = R = 0.50 m. The circular radius and the projectile height are completely different quantities. The radius determines the UCM speed via v = 2πR/T; the height then follows from v²/(2g). Confusing geometric R with kinematic H gives an answer off by a factor of ~25 here.
- 9
Similar NEET-style question
A toy car moves in a horizontal circle of radius 0.80 m completing one revolution every 0.40 s. If the car is flicked vertically upward with its circular-motion speed, find the maximum height. (g = 10 m/s².)
Answer sketch: v = 2π(0.80)/0.40 = 4π m/s; H = (4π)²/(20) = 16π²/20 ≈ 7.9 m.
---
What to remember before solving Uniform Circular Motion questions
Which Uniform Circular Motion formulas do you need for NEET?
1 formula — click to collapse
Centripetal acceleration in uniform circular motion
An object moving in a circle of radius r at constant speed v has acceleration of magnitude v^2/r (or equivalently omega^2 * r) directed toward the centre. This is centripetal (radially inward), not tangential.
| Symbol | Quantity | SI Unit |
|---|---|---|
| a_c | Centripetal acceleration | m/s^2 |
| v | Tangential speed | m/s |
| r | Radius of circle | m |
| omega | Angular speed | rad/s |
Valid when
- Speed v is constant (uniform circular motion)
- r and the centre are well-defined (instantaneous radius of curvature for general curved motion)
Do NOT use when
- Non-uniform circular motion (then there is also a tangential acceleration component)
Where do students lose marks on Uniform Circular Motion?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
3 items — click to collapse
Category: Overthinking
Student uses the radius R as the projectile launch height or fails to compute the UCM speed from period.
When it triggers
Question describes a particle in UCM with given (R, T) then says 'now launched vertically up with same speed; find max height'.
How to avoid
Step 1: speed v = 2πR/T (from UCM). Step 2: max projectile height H = v²/(2g) = (2πR/T)² / (2g). Don't shortcut by setting H = R.
Category: Similar Terms
Student claims velocity is constant in uniform circular motion (it's not — direction changes).
When it triggers
Question asks 'in uniform circular motion at constant speed, which is also constant?'
How to avoid
In UCM: SPEED constant; KE constant. VELOCITY (vector) NOT constant. ACCELERATION (centripetal, magnitude v²/r) constant in MAGNITUDE but NOT in direction.
Root cause: concept gap
Correction
Acceleration is the rate of change of VELOCITY (a vector), not speed. In uniform circular motion, the speed is constant but the velocity direction changes continuously, giving a centripetal acceleration of magnitude v^2/r toward the centre.
Wrong option pattern
Distractor option says 'a = 0 because speed is constant'.
More in Kinematics: 11 exam traps and mistakes · 5 formulas · 8 question patterns from its other lessons.
Uniform Circular Motion questions from past NEET papers
1 question from NEET 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
A particle moving with uniform speed in a circular path maintains:
1Constant velocity
2Constant acceleration
3Constant velocity but varying acceleration
4Varying velocity and varying acceleration
NTA Answer: Option 4(revised_final)
How does NEET ask about Uniform Circular Motion?
2 recurring patterns from past papers — click to collapse
A particle in uniform circular motion (period T, radius R) is subsequently launched vertically upward with the SAME speed (v = 2πR/T); find max height. H = v²/(2g) = (2πR/T)²/(2g).
Multi StepMedium
Common distractors
uses radius as height
Surface confusion of geometric R and projectile H
Particle in uniform circular motion at constant speed; identify which of speed / velocity / acceleration / kinetic energy is constant. Speed and KE constant; velocity (vector) and acceleration (centripetal) are NOT constant — direction changes.
InterpretationEasy
Common distractors
treats velocity as scalar
Conflates speed (scalar) with velocity (vector)
Sources
NCERT refs: Class 11 Physics Chapter 3, p.42
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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